I've tried to prove a property of a kind of iterated multivariable functions I am playing with and I thought I did it, but now I've found a lot of counter examples.
I'm still going to school, but I hope the formalism isn't to horrible.
I started by defining injectivity, surjectivity and bijectivity with respect to a single variable for multivariable functions:
Def: A funtion $f(x,y):\mathbb{R}²\to\mathbb{R}$ is injective/surjective/bijective with respect to $x$ if and only if $f_y(x)=f(x,y)$ is injective/surjective/bijective for all $y\in\mathbb{R}$.
I thought that I proved the Proposition that $f(x_0,y_0)=x_0$ was true if $f(f(x_0,y_0),y_0)=x_0$ is true and $f$ is bijective w.r.t. x and continuous, but I found some counter examples.
One of them is $f(x,y)=\cos(y)-x³$. I wanted to illustrate this with some pictures, but sadly I haven't gotten any reputation yet, so I can't post any.
First I "proved" a Lemma:
If the the system of equations $$f(x)=y\\f(y)=x$$ has any solutions and $f$ is continuous, x=y is true for at least one of them (this seems kind of obvious, but I couldn't find a proof online).
Proof by contradiction:
assume $f(x)\neq x$ for all $x\in\mathbb{R}$.
It follows from continuity that $f(x)<x$ for all $x\in\mathbb{R}$ or $f(x)>x$ for all $x\in\mathbb{R}$.
For $f(x)<x$ this means that $f(f(x))<f(x)<x$,
but the original system of equations means that $f(f(x))=x$.
This is a contradiction.
The proof is exactly the same for the case $f(x)>x$.
this proves the Lemma.
Now I tried to prove the Proposition.
Proof:
Assume $f$ is bijective w.r.t $x$ and continuous and $f(f(x_0,y_0),y_0)=x_0$.
It follows from bijectivity that there is exactly one $x':=f(x_0,y_0)$ that solves this with $f(x',y_0)=x_0$.
Fix $y_0$ and set $f_y(x)=f(x,y_0)$.
Then we have $f_y(x_0)=x'$ and $f_y(x')=x_0$.
It follows from the Lemma that $x_0=x'$ for at least one solution, but since $f_y$ is bijective this is the only solution.
This means that $f(x_0, y_0)=x_0.$
q.e.d.
But as I said earlier there are some counter examples, though it does seem like $f(x_0, y_0)=x_0$ is really true when $f$ is bijective w.r.t. $x$ and continuous and $f(f(f(x_0,y_0),y_0),y_0)=x_0$, though I haven't proved that yet.
Thank you for your help in advance.
What you proved in your lemma is that if there exist $x, y \in \mathbb{R}$ such that $f(x) = y$ and $f(y) = x$ for a continuous function $f: \mathbb{R} \to \mathbb{R}$, then there exists $z \in \mathbb{R}$ such that $f(z) = z.$ It is in general not true that $z = x = y$, because e.g. for $f(x) := -x$ we have $f(1) = -1$ and $f(-1) = 1$, but we only have $f(0) = 0$.
In your proof of the proposition you have that $f_y(x_0) = x'$ and $f_y(x') = x_0$, so you can apply your lemma and deduce that there exists $z \in \mathbb{R}$ such that $f_y(z)=z$. But you can not infer from your lemma that this $z = x_0 = x'.$ This is the (or at least one) error in your proof.
What is in my opinion worthwhile to think about is why this error happened. I reckon this is because you phrased the existence of a $z \in \mathbb{R}$ with $f(z) = z$ in your lemma as "$x=y$ is true for at least one of them", which is not a particularly precise mathematical statement. I recommend taking a look at how to quantify variables using logical quantifiers.